Equal sets and Subsets

Solving \(x^2-4=0\) gives \((x-2)(x+2)=0\), so the integer solutions are \(x=2\) and \(x=-2\). However, the definition of \(A\) asks specifically for a natural-number solution. Under the usual school convention, 2 is natural but −2 is not. Hence \(A=\{2\}\), and since \(B=\{2\}\), the two sets are equal. Therefore option A is correct.

The positive divisors of 24 are 1, 2, 3, 4, 6, 8, 12, and 24. Selecting only the even divisors gives \(A=\{2,4,6,8,12,24\}\). Every element of \(\{2,6,8\}\) belongs to this set, so it is a subset of \(A\). Each other option contains at least one element that is not in \(A\): 1 or 3, 5, or 9. Therefore option A is the only correct answer.

Every set is a subset of itself, so \(A\subseteq A\). However, a proper subset must be strictly smaller than the original set and is usually written with \(\subset\) or \(\subsetneq\). The set \(\{2,4,6\}\) is exactly equal to \(A\), so it is a subset but not a proper subset. The other three choices are proper subsets because they contain fewer elements than \(A\). Therefore option B is correct.

The prime factorization of 30 is \(30=2\times3\times5\). Therefore its prime divisors are exactly 2, 3, and 5, so \(A=\{2,3,5\}\). This is the same set as \(B\), which means \(A=B\), not merely a proper subset. The numbers 1 and 30 are not prime numbers, so option D is also incorrect. Hence option A gives the correct relation.

To form \(C\), select from \(A\) only those elements that are even. The even elements of \(\{1,2,3,4,5,6\}\) are 2, 4, and 6, so \(C=\{2,4,6\}\). This is exactly the set \(B\), hence \(B=C\). Since 2, 4, and 6 belong to \(A\), and A also contains 1, 3, and 5, B is a proper subset of A. Therefore option A is correct.

The statement A ⊆ B says that every element of A belongs to B. The additional statement that B has no element outside A says that every element of B belongs to A, or B ⊆ A. Thus both inclusions hold: A ⊆ B and B ⊆ A. By the criterion for equality of sets, A = B. The value n(A) = 8 is consistent with this and also implies n(B) = 8, not 16. Hence option A is certain.

Solve the absolute-value inequality: |x − 2| < 2 gives −2 < x − 2 < 2. Adding 2 throughout yields 0 < x < 4. Since x must be an integer, A = {1, 2, 3}. The set {1, 3} contains only elements of A, so it is a subset of A, but it does not contain 2 and therefore is not equal to A. Hence it is a proper subset. Option A equals A, while C and D contain elements not belonging to A.

The condition x ∈ N and x divides 18 means that A contains all positive divisors of 18. These divisors are 1, 2, 3, 6, 9, and 18, so A = {1, 2, 3, 6, 9, 18}. This list is exactly B; hence A = B. Option B is false because 1, 6, 9, and 18 are not prime. Option C is false because the two sets have all the same elements. Option D is false because 18 divides itself.

The condition x ∈ Z means that only integers are allowed. The inequality -3 ≤ x includes -3, while x < 2 excludes 2. Listing all integers between these limits gives -3, -2, -1, 0, and 1. Thus A = {-3, -2, -1, 0, 1}, making option A correct. Option C incorrectly includes 2, and option D omits -3.

The phrase x < 10 restricts x to the natural numbers less than 10. Under the convention N = {1, 2, 3, ...}, A = {1, 2, 3, 4, 5, 6, 7, 8, 9}; these are exactly the one-digit natural numbers, so A = B. Even if a text includes 0 in N, both definitions include the same one-digit natural numbers, so their equality remains true. Option D is not universally correct because it depends on whether 0 is included in N.

For an integer x, the inequality |x| ≤ 2 means that x lies between -2 and 2 inclusive. The integers satisfying this condition are -2, -1, 0, 1, and 2. Hence A = {-2, -1, 0, 1, 2}, which is exactly the given set B. Therefore A = B. Option B omits the negative integers, option D omits the values between the endpoints, and option C is false because the two sets are equal rather than properly contained.

An element of A must satisfy both conditions: it must be even and it must be prime, while also being less than 10. The only even prime number is 2, and 2 is less than 10. The numbers 4, 6, and 8 are even but composite; 3, 5, and 7 are prime but odd. Therefore A contains only 2, so A = {2}. Option A is correct.

The set A contains three elements: the number 1, the number 2, and the set {1, 2}. The inner set {1, 2} is counted as one single element of A, not as two additional elements. Therefore, n(A) = 3. This illustrates that a set itself can be an element of another set, and elements must be counted according to how they are listed at the outermost level.

To determine A, solve x^2 = 9. The integer solutions are x = -3 and x = 3, so A = {-3, 3}. Set B contains these two elements and one additional element, 9: B = {-3, 3, 9}. Hence every element of A belongs to B, but A and B are not equal. Therefore, A is a proper subset of B.

For set A, solve x² − 1 = 0. Factoring gives (x − 1)(x + 1) = 0, so x = 1 or x = −1; both are integers. Thus A = {−1, 1}. For set B, the integers strictly between −2 and 2 are −1, 0, and 1. The condition x ≠ 0 removes 0, leaving B = {−1, 1}. Since A and B contain exactly the same elements, A = B. Therefore, option A is correct; B, C, and D are false.

Set B has exactly three elements: 2, 3, and 5. A subset with three elements must therefore contain all elements of B, so the only such subset is B itself, namely {2, 3, 5}. Since every element of B is also an element of A, B is a subset of A; in fact, it is a proper subset because A contains additional elements 1, 4, and 6.

Assuming N denotes the positive natural numbers, A = {1, 2, 3, 4, 5, 6}. The positive divisors of 6 are B = {1, 2, 3, 6}. Every element of B is in A, so B is a subset of A. However, A also contains 4 and 5, which are not in B. Thus B is not equal to A; it is a proper subset of A, making option B correct.

Solve the defining equation: x² − 2x = x(x − 2) = 0. Therefore, x = 0 or x = 2. Both values are integers, so A = {0,2}. Since B is also defined as {0,2}, the two sets contain exactly the same elements and therefore A = B. Option B omits 0, option C incorrectly claims a proper subset relationship, and option D ignores the two valid solutions.

A prime number is a natural number greater than 1 that has exactly two positive divisors: 1 and itself. The prime numbers from 1 through 20 are 2, 3, 5, 7, 11, 13, 17, and 19. Number 1 is not prime because it has only one positive divisor. Thus A and B contain exactly the same elements, so A = B. Option D incorrectly includes 1.

The positive multiples of 5 are 5, 10, 15, 20, 25, 30, and so on. The strict condition x < 30 excludes 30, leaving exactly 5, 10, 15, 20, and 25. These are precisely the elements listed in B. Therefore A and B contain the same elements, so A = B. Option B wrongly includes the boundary value 30; options C and D do not describe the actual relationship.